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I am currently doing a question to get all possible eigenvectors of a matrix, and while I believe I got the correct answer (and WolframAlpha also says that my answer is correct), I was marked wrong. I just want to know if I made some mistake, or perhaps the answer that the online system I had was wrong. The question is as follows:

Find all distinct (real or complex) eigenvalues of A. Then find the basic eigenvectors of A corresponding to each eigenvalue. For each eigenvalue, specify the number of basic eigenvectors corresponding to that eigenvalue, then enter the eigenvalue followed by the basic eigenvectors corresponding to that eigenvalue.

$$ A = \begin{bmatrix} -6 & 5 \\ -10 & 8 \end{bmatrix} $$

I was able to determine the eigenvalues to be $1+i$ as well as $1-i$, which it did accept, except I got (as well as WolframAlpha) the Eigenvectors below:

$$ 1+i: \begin{bmatrix} \frac{7}{10}+\frac{i}{10} \\ 1 \end{bmatrix} $$ $$ 1-i: \begin{bmatrix} \frac{7}{10}-\frac{i}{10} \\ 1 \end{bmatrix} $$

While the correct answer it gave was as follows:

$$ 1+i: \begin{bmatrix} -2+i \\ -3+i \end{bmatrix} $$ $$ 1-i: \begin{bmatrix} -2-i\\ -3-i \end{bmatrix} $$

Would anyone be able to tell me if I am correct and there is some issue with my homework system, or I am wrong and I should be quadruple checking my work? This has been driving me crazy all night.

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Hint:

$$ \left(\frac{7}{10}+\frac{i}{10}\right)(-3-i)=-2-i $$ do the same for the other eigenvector. $$ \left(\frac{7}{10}-\frac{i}{10}\right)(-3+i)=-2+i $$

It seems that you have changed the eigenvectors.( see http://www.wolframalpha.com/input/?i=eigenvalues+%7B%7B-6%2C5%7D%2C%7B-10%2C8%7D%7D)

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  • $\begingroup$ I do see that the correct eigenvectors are multiples of my vectors, but I still don't see why. That WolframAlpha link you showed still shows the answers I gave as the eigenvectors for that matrix. $\endgroup$ – Nick Taylor Dec 6 '15 at 17:02
  • $\begingroup$ Eigenvectors are not unique ! To any eigenvalue is connected an eigenspace, i.e. a linear space spanned by the eigenvectors, so if $v$ is an eigenvector also $\alpha v$ is an eigenvector of the same eigenspace, connected to the same eigenvalue. $\endgroup$ – Emilio Novati Dec 6 '15 at 17:37
  • $\begingroup$ Ah, that clears some things up. I tried that question again and got a different matrix under which the process I used did work, so it must have just been that particular question was not properly processing my answer. Thanks ! $\endgroup$ – Nick Taylor Dec 6 '15 at 17:40

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